D&we Applied Mathematics l(l979) 229-252 @ Nti-Holland Publishhag Company

CONDORCET PROPORWONS AND KELLY’S CONJECTUWW

Peter C. FISHBURN

Bell kabmtw& Mumy Hill, NJ 075V4, U.S.A:

William V. GEHRLEIN Uniucrsity of Delawan, Nmtwk, DE, C’.S.A.

Eric MASKIN

&~~acbt?~ fnst&e of Techo~gy, Cambridge, MA, U.S.A.

Received 12 July 1979

Let C(m, n) be the proportion of all n-tuples of liners orders on a set of m alternatives such that some altematke x is ranked ahead of y in at Icast &z of the orders, for each y # x. Kelly proved~atC(m,n)cC(on,r,+l)form~3andoddn~3,andthatC(m,n)>C(on,n+l)for mP3andevenna2.Heal~~~~thatC(na,n)>C(m+l,n)form~3andn=3or na5, and that C(m,n)>C(m,n+2) for ma3 and n=l or n~53. The first of these conjechues is shown to be tnze for n = 3, and for m = 3 and odd n. The second conjecture is established for m ~{3,4} and odd n, and for m = 3 and all large even n.

1. lnbnrdpctiob

A p#ik is a f&e list of linear orders on a finite set of alternatives. A profile is a Conhrcet profile if and only if it has a Condorcet alternative, which is an

a3temative x such that, for every alternative y # x, at least as many orders in the list have x ranked ahead of y as have y ranked ahead of x. 111 other words, a Condorcet alternative is a simple majority alternative. The Condorcet propotion C(m, n) for n-term profiles on M alternatives is the number of n-term Condorcet profiles on m alternatives divided by (m!)“, the total number of n-term profiles on m alternatives.

Most of what is presently known about Condorcet proportions can be found in Garman and Kamien [SJ, Niemi and Weisberg 1121, DeMeyer and Plott [3j, May [ 111, Kelly [lo], and Gehrlein and Fishbum 16, 71. All but one of these deal with precise computations and approximations of C(m, n). The exception is Kelly, who examined trends in C(m, n) and showed that

C(m,n)<C(m,n+l) for ma3 and odd na3,

C(m,n)>C(m,n+l) for ma3 and even ~122.

1 The work of Fishbum and Gehrlein was partly supported by Grant No. SOC 77-22941 from the National science Foundation to The Pennsylvania State University.

229

230 P.C. Fishbum, W. V’. Geld&, E. Maskin

Kelly also proposed:

Conje&ue 1. C(m, n)>C(m+l, n) for pna3, n=3 01 ?tH.

ConjW2. C(m,n)>C(m,n+2) fOrmHJ,n=lW ?IE:3.

He noted that “proofs of these two conjectures would be important contt%utions to the formal voting literature” and discussed some of their difficulties. With r~ as the number of voters, Conjecture’ 1 says that, for fixed n ~{3,5,6,. l .}, the Condorcei proportions decrease in m B3; Conjecture 2 says that, for fixed ~lt E (3,4, . . .}, the Condorcet proportion for n voters exceeds the proportion for n +2 voters, except when n = 2 since C(3,2) = C(3,4) = 1. The main purpose of this paper is to examine Conjecture 1 for n = 3 and

Conjecttire 2 for m = 3. Section 2 will prove that Conjecture 1 is true for three voters. Section 3 proves that Conjecture 2 is true for three alternatives and odd numbers of voters, and Section 5 notes that Conjecture 2 is true for three alternatives and almost all even numbers of voters n 34. In particular, for m = 3 and n even, we shall prove that Conjecture 2 is true for all even n greater than some finite N. Several ancilliary results for m = 3 and n even will be noted in Section 4. Finally, Section 6 comments on the conjectures for odd n and larger values of m.

Although we deal only with several basic cases, the proofs tend o e involved and consume most of the paper. Our experience with these cases s g MS that the more general cases will be extremely difficult to resolve. Nevertl .I s, we would encourage others to try them using either extensions of the proof hniques used here or entirely new techniques that we cannot now foresee.

The proof in the next section is based on the proportions forr ation described above. Thereafter we shall turn to probabilistic proof technique l~ which C(m, n) is interpreted as the probability that there will be a Condorcet itemative for the m-alternative, n-voter situation in which sach voter indepen$ fitly selects one of

. the m! linear orders on the alternatives as his preference on .er according to the equally-likely probability distribution that assigns probability I/m! to each of the m ! linezu= orders.

2. cW#cture 1 for a=3

All profiles in this section will be 3-term, or 3-voter, profiles. We shall prove that C(m, 3) decreases in m 2 2, where clearly C(2,3) = 1.

Theorem 1. C(m, 3) > C(m + 1,3) for all ??a E{2,3,. . .}.

Let B(m) be the number of Condorcet profiles on m alternatives that have a fixed alternative x as the necessarily unique Condorcet alternative. Since

Z(m, 3) = mB(m)/(m!)3, the inequality in Theorem 1 reduces to (m + l)*m.B(m) > B(m+l) for mB2, or

(m+2)2(m+l)B(m+1)M3(m+2) for mal. (1)

We shall approach (1) through a series o four k mas. The following index set wiU be used in the first two lemmas and at the conclusion of the section:

A, ={(Q, Us, 423) E{O, 1,2, . . .}% u1 + u2 + a3 S m}.

PWM& For fixed x in a set of m+l alternatives let a, be the number of alternatives in the set that voter i prefers to (ranks ahead of) x, for i = 1,2,3. Then x can be the Condorcet alternative for a profile only if (a,, a*, U&E &. Given (~1, a2, a3) E &, there are q!(m -q)! linear orders on the alternatives in which exactly cl, alternatives are ranked ahead of x, and there are mi~~aI!a2!a,!(m- UrU2 - a3)!] distinct ways that the three ahead-of-x sets with ai members can be chosen without duplication, which is precisely what is needed for x to be the Condorcet alternative. Therefore B(m + 1) equals the latter expression times the product of the three R!(m- q)!, summed over A,,,. 0

For any m and (aI, u2, ad let

c

F,+&bU2,U3)=(m+1-ad(m+ 1- U*)(U3+ 1)

+(m+l-a,)(m+l-a,)(a2+1)

+(m + 1- a,)(m + 1- &(a, + 1) +(m+l-a,)(m+l-a*)(m+l-a,),

G,,,+l(al, ~12, a3)= m!(m-aJ!(m-a&(m-a3)!/(m-al-a,-a,)!,

with B(m + l)=C, G,,,+l(~I, a2, a,) by Lemma 1.

~smmn 2. B(m + 2) = c\, r”,+I(a,, ~2, RX%,+A~I, ~2, a,).

Proof. Given (a,, a2, a3)e A,,,, Gm+, ( ul, a*, a,) is the number of profiles on m + 1 alternatives in which x beats each of the other m alternatives by a simple majority. When the (m + 2)nd alternative is added to such a profire, it will be beaten by x under simple majority if and only if no more than one voter has the new alternative ranked ahead of x, and Fm+1 (al, a2, a,) is the number of ways this can be done. Therefore B(m +2) equals F,+l(al, a2, a,)G,,,+,(a,, a2, a3) summed over A,,,. 0

232 PC. Fishbw n, W.V. Gehddn, E. l&skin

Lemma 3. If integers a a0 and 5>0 satisfy baa, then

l+a+a(a--l)+ a(a-1)-**1. b+l b 6(6-l) l l *+6(6-l) l l l (6-07j=b-a+1 l

Proof. The conclusion clearly holds if a = 0 or 6 = 1. For u > 0 and 6 B 1, the sum in Lemma 3 equals 1 +(a/b) [sum for a - 1 and 6 - I]. If the lemma% conclusion holds for 6 - 1 and all OS u e b- 1, then the latter [sum] is 6/(6-a + 1), with 1+ (alb)[bl(b - a +- l)] = (6 + 1)/(6 - a + 1). The lemma then follows from induction on 6. Cl

Lemma 4. If integers u1 a 0 and a2 90 satisfv al + u2S m, then m -al-a2

I: F,+hh, a2, aWm+dul, a2, a,) s a3 =o

s (m +212(, + l)m-$-azG,+,(a,, u2, a,), a3=0

with c in the condusion if al + a2 >O.

Proof. Given a, a0, u2 ~0 and a, + a,~ m, the definition of G,,,+r prior to Lemma 2 implies that

m-al -a2

c G,+Aa,, a2, ad =

m!(m - al)!(m - a2)! (m!)

as=0 (nz-al-a*)!

x 1+ 1 m-a,-tr, +-.*+ (m-a,-& l . ’ 1

nr m(m-1) l ’ l (a,+u2+1) 1 = m!(m - a,)!(m - a,)!(m + l)!

(t?l-Ul- a,)!(a,+a2+1) by Lemma 3.

When the first and fourth terms in the def.nition of Fm+l that precedes Lemma 2 are grouped, we get

m-a,-a2

1 Fm+hh, u2, u3G,+Aul,u2,u3)= a3 =o

= 1 (m +2h .f I- a,)(m + 1 - a2)G,,,+Ja1, u2, aS) a3

x(ra + 1 - a,)G,,,(a,, u2, aj)

=im~2)im+l-a,)im+1-a2) m!(m-a,)!(m -a,)!(m + l)!

(m_u,_u ~r(u~+u2+1)

2. *

+[(m+1-a,)(a2+l)+(m+1-~2)(~If1)]

xm!(m-uI)!(m-a,)!(m+l)!(m+2) (m-a,-u,)!(a,+u2+2)

-t und Kelly’s con-

by the result just proved and Lemma 3. The ratio of

233

is therefore

=: a ~~-+2[(a~+a,+2)m2+(3uI+3a,+6)m+f(a,, a,>] 12

where

f(U,, a2)=4+a,+a2-a~-a~-a~a2-a,a~-2aIa2.

Since

(m+2)2(m+l)=a T::, 12

~~(a,+a2+2)m2+(3a,+3a2+6)m+(2aI+2a2+4)],

and since f(a,, aJ~2a,+2u~+4 with < holding if al+a2>0, it follows that

cF'+h a29 a3)G,+dal, a2,a3l"(m +2)2(m+1) 1 Gm+l(a,, a2, as), a3 Q3

with < holding if a,+a,>O. m

Finally, when both sides of the inequality in Lemma 4 are summed over a, and a2 for a+O, a2a0 and a1+a2~m, we get

1 K,+d.al, a2, a3)G,,,+dal, a2, ad < (m + 212(m + 1) 1 G,,+~Q~, a2, =A a, A,

given m 3 1. Then (I) follows immediately from Lemmas 1 and 2.

For convenience in this and the next two sections we shall let C(n) = C(3, n), so that C(n) is the number of n-term Condorcet profiles on three alternatives divided by 6”. Our pupse in the present section is to prove

234 P.C. Fishbum, W.V. Gehrkin, E. bfd~kiit

Tlheorem 2. C(n) > C( n + 2) for all odd n a 1.

We shall prove this using the vernacular of the random-voters model with {1,2,3} as the set of three alternatives. The main part of the proof will be preceded by the following lemma in which, for euen n,

Q,,(t) = Probability {(number of voters who prefer 1 to 3) - (number of voters who prefer 3 to 1) = t 1 n is euen, in voters prefers 1 to 2 and the other fn voters prefer 2 to 1).

By the symmetry of the conditioning event, Q,,(t) = Q&t) for t ~{2,4,. . . , n}.

Lemma 5. If n a 2 is even, tJ~n Q,,(O) > QJ2) > Q,,(4)> 9 l l > Q,,(n).

Proof. Since Q&O) = 5 and Q*(2) = QJ-2) = $, the conclusion holds for n = 2. Assume henceforth that IZ is even n 34 and t E {-n, 9-n +2,. . . , n -2, n}. Although II is fixed in the follovwing, the proof applies to any n as just specified. By dividing the 18 voters into a subset of two voters and a second subset of the other it -2 voters such that,. within each subset the same number of voters prefer 1 to 2 as prefer 2 to 1, it is evident that

Q,(t) =pQ~_l(t-2)+~Qn_2(t)+~Qn_a(t+2).

Each Qla-* term on the right-hand side can be decomposed in like manner in terms of Q,,+ and Qz (which provides the 3 and $ multipliers). In general, for each positive integer k for which II -2k 3 2, there are integers fk(a) for a E @,VL.., 2k) such that

9kQnW = fdWQ,,--2dO + i f&j)[Q,,-df -2j)+ Qn-& +2j)] j-1

with

fk(O)WXW* l l ~f,cwxh

The previous expression for Q,,(t) in terms of Q,,_* shows that this holds for k = 1 with fl(0) = 5 and f,(2) = 2. This completes the proof for n = 4. When n a6, induction on k then shows that it is generally true: assuming its truth for all k up to some k for which n -2k 24, the breakdown of each Q,,_;Zk in terms of Q n-2k-2 = Qn_2(k+1j shows that

fk,l(a)=2fk(a_2)+5f,(a)+Zfk(a+2)

condorrwt pmpodns and Kdy’s con- 235

for each a ~{0,2 , . . . ,2(k + 1)) where, by convention, &(a) = 0 for Q Sk

and fkO=fkOf then for 42~{0,2,. e l 92(k+1)}, fk+l(~)>fk+l(~+2) ti 2~((r_2)+3f,(~)>3~~(~92)+2~~(~+4), which is true by the induction hypothesis.

Sk Q&) # 0 iff a EC-2,0, 2}, the conclusion just proved shows that, when k =&n-2) and te{O,2,. . .,n},

gkQ~(t)=fk(t-2)Q*(2)+fk(t)Q*(0)+fk(t+2)Q2(-2)

=;Sfk(r -2)+8fk(t)+gfk(f+2),

with fk(n+2)=0 and fk(-2)=fk(2). Hence, when k=&n-2) and TV (0 2 9 l l l 9 n-21, Qn(tPQ,,(t+2) iff 2fk(t-2)+3fk(t)>3fk(t+2)+2fk(t+4), wiich is true since fk(0)>fk(2)>a l >fk(n-2)>0. 0

we now prove Theorem 2. Let PI(n) be the probability that alternative 1 is the Condorcet alternative when there are n 3 1 voters and n is odd. Since C? is odd, symmetry implies that C(n)=3P,(n). Hence to prove Theorem 2 we shall show that

PI(~)M5(n+2) for all odd nal.

since P,(l)=f>P,(3) =& this is true for n = 1. We assume henceforth that na3.

Given n voters with PI odd, let (ar, 0) denote the event in which ii) CY of the dtematives in {2,3} are beaten by alternative 1 by exactly one vote (e.g., &z + 1) voters prefer 1 to 2 and the other #n - 1) perfer 2 to 1), (ii) p alternatives in {2,3} beat alternative 1 by exactly one vote, and (iii) the other 2 - (CU + p) alternatives in {2,3} are beaten by 1 by three or more votes. Let P(~M, p) be the probability that (% /3) obtains. Then

In addition, when two new voters are added to the n votes who are represenied in (at, @, we see that

P~(n+2)=P(O,O)+~P(1,O)+~P(2,0)

+$P(O, l)+$P(O, 2j+$P(l, 1).

For example, if (a, 8) = (2, G) obtains for the first n voters, then alternative 12 .wiU be the Condorcet alternative for the 1c +2 voters iff at least one of the twa new voters prefers 1 to 2 and at least o’ne of the two new voters prefers 1 to 3. The probability of the latter joint event for the two new voters is s.

236 P.C. FIshbum, W.V. Gehrlein, E. Maskin

According to the preceding paragraph, PI(n) >P &I +2) iff

9P(1,0)+14P(2,0)>9P(O, 1)+4P(O,2)+8P(l, 1). I 1

To verify the latter inequality we use a conditional analysis of the P(ar, 8) for the n original voters as follows. Let A and A’ be respectively the event that 1 beats 2 by one vote (for the n voters) and the event that 2 beats 1 by one vote. By symmetry, P(A) =P(A’), where P( ) denotes the probability of the event in parentheses. Conditional analysis and symmetry considerations yield

P(l, 0) = 2P(l beats 3 by 33 votes i I)P(A),

P(2,O) = P(l beats 3 by 1 vote 1 A)P(A),

P(0, 1) = 2P(l beats 3 by 23 votes 1 A’)P(A)

= 2P(2 beats 3 by a3 votes 1 A)P(A),

P(O,2) = P(2,0),

P(1) 1) = 2P(3 beats 1 by 1 vote 1 A)P(A).

When these are substituted into the preceding inequality and P(A) >O is canceled, we see that Pl(n)>P,(n +2) if

9P( 1 beats 3 by 23 votes 1 A) + 5P( 1 beats 3 by 1 vote 1 A)>

> 9P(2 beats 3 by a3 votes I A)+SP(3 beats 1 by 1 vote I A).

Since these conditional probabilities are not afkcted by the. specific manner in which A is realized - i.e., which voters prefer 1 to 2 and 2 to 1 -we fix one voter who prefers 1 to 2 and let the remaining n - 1 voters be evenly divided between 1 versus 2. Then, with Q,_I(t) referring to the latter n - 1 voters, it follows from the definition of Q,,-I(t) that

P(1 beats 3 by 23 votes 1 A) = $Q,-t(2) + 1 Q,,_I(t), 1=4

P(l beats 3 by 1 vote 1 A) = iQ,_,(0)+$Q,,_1(2),

P(2 beats 3 by 33 votes I A) =$Q,,_,(2)+ y Q,,_I(r), IS”4

P(3 beats 1 by 1 vote 1 A) =$a,,-,(2)+fQ,_,(O).

The final equation here uses the fact that Q,_l( -2) = Q,,-l(2), and the third equation uses symmetry as fohows: (2 beats 3 by 33 votes 1 A) iff (2 beats 3 by 2 votes in the n - 1 (whose probability, by symmetry, is the same as 1 beating 3 by 2 votes, i.e., Q,,_.1(2)) and the fixed voter prefers 2 to 3 (with probability 4)) or (2

beats 3 by 4 or more votes (with probability Q,&4)+ Q,,_,(6)+ - l l by symmetry)}. When the four preceding equalities are substituted into the former inequality for &(n)>&(n +2), we get &(n)>P,(n +2) iff Q,+(O)> Q,,_&), which is true by Lemma 5 for odd n ~3.

The obvious fact that differentiates even n from odd n is that a Condorcet profile for even n can have more than one Condorcet alternative. Moreover, even when an even-n profile has a unique Condor& alternative, another alternative can tie the Condorcet alternative so long as it is beaten by something else. For example, we can have 1 and 2 tied with 1 beating 3 and 3 beating 2. Then I is the unique Caldorcet alternative although it does not have strict majorities over both of the other alterYratives.

This section wiU focus on a specifk alternative either as the unique Condor& alternative that beats the others or as one of the possibly more than one Condor& alternatives. Kelly’s Conjecture 2 for three alternatives and even n will be examined in the next section. According to our present concerns, let { 1,2,3} be the set of three alternatives and let

PI(n) = Probability (alternative 1 is a Condorcet alternative and neither 2 nor 3 ties 11 n voters),

?‘&I) = Probability (alternative 1 is a Condorcet alternative 1 n voters).

If n is odd, then Tl(n) =&(n), but if n is even, then T,(n)>&(n) for n&L

Tw 3. P,(n)<~Jn+2) for ~ZZ even na2.

Theorem 4. T,(o;t)>T,(n+2) for all even ~132.

Theorems 2 through 4 show that TI for even n behaves in the same way as P, for odd n, while PI for even n reverses the behavior of PI for odd n. Since ties for the even-n case have probability zero in the limit of II, the limit probability P&+-which is one-third of Guilbaud’s number [S, 91 and is approximately 0.30408- is approached from above by PI for odd n and TI for even n but is approached from below by PI for even n.

Before proving Theorems 3 and 4 we shall establish a lemma that will be used in the proof of Theorem 3.

Lemma 6. n[Q,,_,(@) - Q,,-,(4)] = 9Q,,(2) for all even n 24.

Proof. Let 2r = n -- 2 and partition the 2r voters who are evenly divided on 1 versus 2 ,in the conditioning event for Q&t) into r groups of two, voters ,each such that, within each group, one voter prefers 1 to 2 and the other prefers 2 to 1. Then

n[Q,,-z(O) - a,-,(4)5/2 = (r + 1>tQdI) - Q2,(4>3

t1+m1

-c

rf22k-2y-2k+2

k=2 k!(k -2)!(r-2k +2)!9’ ’

which follows from the within-group probabilities of #, 8 and 3 respectively that both voters prefer 1 to 3, both prefer 3 to 1, and the two are divided on 1 vs. 3. When the k = 0 term is separated from the first sum and k is replaced by k + 1 throughout the second sum, the preceding expression can be written as

Cr121

c

,.r22ky-2k

- &el (k+l)!(L)!(r-2k)!Y

r!p5’-*k

k=l k!(k + l)!(r-2k)!9’

9 ‘d2’ (t + 1)!22k+L5’-Zk = YQ2tr+lb(2)

=--

2

z

k-0 k!(k + 1)!(~-2k)!9~“’ .

. Therefore n[Q,,-2(0) - Q,,_,(4)] = 9QJ2). 0

Proof of Theorem 3. This proof is similar to the proof of Theorem 2 but is slightly more complex for reasons that will become apparent. Throughout the proof, n is even and n B 2. In this context we let (cq 0) denote the event in which (i) (Y alternatives in {2.,3} are beaten by alternative 1 by exactly wo votes, (ii) /3 alternatives in {2,3} tie alternative 1 (e.g., $n voters prefer to 2 and the other@ prefer 2 to l), and (iii) the other 2 -(cu + 0) alternatives in {2,3} are beaten by alternative 1 by four or more votes. With P(cu, @) the probability of event ((w, 0) for n voters, the addition of two new voters to the n voters represented in (a, /3) gives

P~(n+~)-P~(n)=[P(O,O)+~P(1,O)+~gP(2,0)

+$P(O, l)+$P(O, 2)+$P(l, l)]

-[P(O, O)+P(1,O)+P(2, O)].

It foIiows from this and symmetry considerations that PJn +2) > PI(n) iff

condorect pqodons and K&y ‘s conjectures 239

where, n voters,

pl = P(Z ties 2, ‘1 beats 3 by ~4 votes),

pz =P(l ties 2, .l bea’s 3 by 2 votes),

p3 =,P(l ties 2, 1 tie 3),

p4 = P( 1 beats 2 by 2 votes; 1 beats 3 by 24 votes),

ps = P(1 beats 2 by 2 votes, 1 beats 3 by 2 votes).

Let A and B be respectively the event that 1 ties 2 and the event that I beats 2 by 2 votes. Then

p1 = P( 1 beats 3 by r4 votes 1 A)P(A) = pip(A),

pz=P(l beats 3 by 2 votes 1 A)P(A) =&P(A),

p3 = P(l ties 3 1 A)P(A) = &P(A),

p4 =P(l beats 3 by 34 votes 1 B)P(.B) = p&P(B),

ps = P(1 beats 3 by 2 votes 1 B)P(B) = p&P(B),

where pi through pi are defined in context. Since P(A)=2’“(,‘;,) and P(B) = 2-“(nn+l), P(A)/PW = 1+2/n. This fact and substitution of pi through p; into the preceding inequality for Pl(n + 2) > PI(n) yields PI( n + 2) > P1( n) ifI

I

It is easily checked that this hoLs for n = 2, so we assume henceforth that II B 4. By viewing A as composed of two voters who are divided on 1 versus 2 plus the remaining n -2 voters who are evenly split between 1 preferred to 2 and 2 preferred to 1, it follows that

D I I pi =W~-J2)+$C?n-,C4~ + QR-2(36),

~5 = ik.dO) +dQ,,&) +$Q,,-&% t

p;‘~Qn-2(2)+$Qn-*(O), !

where Q,,_@6) = Q,&6)+Qn _&I)+- l l , and where O,,_,(-2) = Q&2) has been used in pi. Similarly, by viewing B as composed of two voters INho both prefer 1 to 2 plus n -2 voters who are evenly split on 1 versus 2,

p;=bQn-2(2)+3Qn-*(4)+Qn-2(~6),

p; =tQ,,(Ol +$Q,&) + bQ,-,C4>.

Substitution into the preceding inequality gives P,(n +2)> P;(n) ifl

(1+21n)[26Q,_2(0)+66Q,,(2)+79Q,_,(4)+818,-,(~6)1>

>[28Q”-*(0)+64Q~-2(23+79Q,-,(4)+81Q,-,(96)~~

240 EC. Fishbum, W.V. Gehdein, 33. Maskin

Since the sum of the Q,,-&) for even t equals 1, and since Qn.J-t) = Q&t) by the line before Lemma 5, we have

a,-&4 = $ - $Q,&O) - Q,p4(2) - Q&4).

When this is substituted into the preceding inequality and fractions are cleared, we get P&n +2)=+,(n) iff

8 1’ 2nCQAO) - Q,,-,(2)]+ 29Q,,-*(O) + 3OQ,,_,(2) +4Q,&4).

Since Lemmas 5 and 6 imply that 9QJ2) > n[Q,,-a(0)-Q,_2(2)], it follows that the right-hand side of the ~preceding inequality cannot exceed 2(9) + 30 = 48. 0

Proof of l’%eorem 4. In the T1 context let (ar, p) be the event for n voters in which (i) CY alternatives in {2,3} tie alternative 1, (ii) @ alternatives in {2,3} beat alternative 1 by exactly two votes, and (iii) the other 2-(ar + 6) alternatives in (2,3} are beaten by alternative 1 by two or more votes. Then, proceeding as in previous proofs with P(cu, p) the probability of (cw, @),

T,(n)-T,(n+2)=[P(O,O)+P(1,O)+P(2,0)]-[P(O,O)+~~(1,O)

+##(2,O)+,‘P(O, l)+$P(O, 2)+$P(l, l)].

Let A and B be respecti\*ely the event that 1 ties 2 and the event that 2 by 2 votes, with respect to n voters. Then

P( 1,O) = 2P( 1 beats 3 by 32 votes 1 A)P(A) = 2q,P(A),

P(2,O) = P(i ties 3 1 A)P(A) =@(A),

P(0, 1) = ZP(1 beats 3 by 32 votes 1 @P(B) = 2q#(B),

P( 1,l) = 2P( 1 ties 3 i D)P(B) = 2q,P( B),

P(O,2) = PU beats 1 by 2 votes 1 B)P(B) = qsP(B),

where q1 through q5 are defined in context. Since P(A) =(&)2’” and P(A)izr’(iz + 2), it follows that

beats 1

.

P(B) =

&(n)- T&t +2) = \rr/2/- 18(n +2) [(gql+7qz)(u +2)-(9qa+8qea+2q&il.

Since this is easily seer1 to be positive fVy: n = 2, n 2 4 will be presumed hence- forth. Then, since reasoning that follows that used to break down ai through pi in

the preceding proof shows that

and since Q,,&t) = Q,,_*(t), it follows that

T&t)-T&+2)= C~C2Q,,-20 - Q,,-&) - CL-201

+[53Q,,-,(o)+91Q,,_,(2)+81Q,_,(~4)~.

Since Lemma 5 implies that the Grst term in brackets is positive, and since. the se&d term in brackets is positive too, it follows that T*(n) > ‘I&z + 2) for all even n. 0

Note on T3. The preceding expression for Tl( n) - T,(n + 2) implies a limit result that will be used in the next section. Here and later we shall let f(n) C- g(n) mean that If(n) - g(n)1 ---, 0 as n --)a. In addition, we define K, by .

( 1 n 2-n

K,= n12 . (n+2) l

Leubma 7. 3[T&t)-T,(n+2)ylK/q.

prwb. When Q,,_@4) =#-HO,_,- Qn_a(2) is used in the final equation of the proof of Theorem 4, we get

3[T,(n)- TI(~ +2)1/K, = %+&@Q,,-&) - a,-&) - Q,,-2(4)1

+2”7[12.5Q,,2(0) + lOQ,,_,(2)].

In view of Lemmas 5 and 6, all terms on the right hand side gq> to zero as n gets

large except for 3. 13

5. ConjeCanre2for m=3andeven n

In this section we shall prove that Kelly’s second conjecture holds for almost all even n ~4 when there are three alternatives. Throughout, the set of alternatives is {1,2,3} with C(n) = C(3, n).

242 PC Rshbu~ WX Cchriein, E. Maskfn

Themem 5. C(n) > C(n ,+ 2) for a/! even n greater than some positive integer N.

In addition, C(n) > C(n + 2) can be established for small even n 24 by direct calculation. For example, C(4) = 1 and C(6) = 1291112%. Although we do not presently have a proof of C( ss) > C(n + 2) for a!! even n B 4, we have no reason to doubt its vahdity.

Aloq with T&I) from the preceding section let

T,,(n) = Probabiiitf (&ematives 1 and 2 are Condorcet alternatives 1 n voters),

T&VI) = Probability (alternatives 1, 2 and 3 are Condorcet alternatives 1 n voters).

Since synunem among alternatives implies that

C(n) = 3T,(n) - 3T12(nj + Tdn),

C(n)-C(n+2)=3[T,(n)-T,(n+2)]-(3[T,,(n)-T&+2)]

+ ITm(n + 2) -- LWII- .

The limiting behavior of 3[ T&I) - Tl(n + 2)] is known lrom Lemma 7. We shall therefore focus on the term in braces in the preceding expression. Let

QJ tI, f2) = Probability ((number of voters who prefer 1 to 3) -(number who prefer 3 to 1) = tl, (number of voters who prefer 2 to 3)-(number who prefer 3 to 2) = t2 1 n voters , in of whom prefer 1 to 2 and $n of whom prefer 2 to 1).

In ali cases, tI and f2 are even integers and n is a positive even integer. By symmetry, Q,&, t2) = Q,, (f*, tI) = Q,, (-tl, -tz). For notational convenience we shall let Q,,( tl, 3 f2) = Q,,( tI, f2) + QJ tl, t2 + 2) + l 9 . with similar conventions for Q,A%, f2) and Q,,Ptl, Ws

To analyze the part of, C(n) - C(n + 2) that involves Tla and Tlz3 we observe first that

T&) = ( ) n;2 2-“Q,(W W,

TI&) = ( n72)2-nQJ0, 0),

T**( n + 2) = 2 -” -2Q”+2( a0, 20)

T,,& + 2) = ( n;2)2-n(3Qn+2@, or.

A straightforward conditional analysis on two voters who are evenly divided on 1 versus 2 shows that

Q~+2(~0,)0)~~9Qn(~2*82) +7Q,,WW +7Q,,(O,a2)

+6Q,,(O, 0)+2Qn(*2, -2)+2Q,((-2,32)

+2Q,,(O,-2)+2Q,(-2,0)+QA-2,.2)]

=&9Q,,(*2, a2)+ Q,(2,2)+ 14Q,,(O, a2)

+6Q,,(O, O)+dQ,((-2, WI-

In like manner,

Q,,+#, 0) =&2Q,(2,2)+4Q,,(O, 2)+3Q,(O, WI-

using

Q,(b2, a2) = Q&O, aO)-2Q,(O, ~0)+ Q,,(O, 0)

and Q”(O, *2) = Q,,(O, SO)- QJO, 0), it follows that

{3CT~*(n)-'P;*(n+2)1+[T~~(n+2)-~~u(n)l)l~=

=3Q~(~O,a0)-~(9n+18)Q,(O,0)-(4n+4)Q,,(O, 2)

+ (n + l)Q,(2,31

+&a + l)[Q,@, aO)- a,(-2, aO)]=

The limiting behaviors of the three parts of this expression are covered by the fallowing three lemmas. As before, f(n)- g(n) means that If(n)-g(n)1 + 0 as n --+ 00. In addition, we shall write f(n)% g(n) if and only if either f(n) c g(n) for all large even n or else f(n)- g(n), and f(n)B g(n) if and only if g(n)9 f(n).

Lemma 8. Q,,(aO, so)-&.

Leaunor 9. nQ&, t,)-341(2n) for tied (tl, tz).

LB 10. (n + l)[QJO, ~0).Q,,(-2, aO)]s 9/(4r)=

The proofs of these lemmas will be deferred until we see how they are used in proving Theorem 5.

Pmof of Thonmn 5. Lemmas 7 through 10 along with Q& tz) m 0 show that,

244 P.C. Rishburn, W. V. Oehrkin, E. Ma&in

since [C(n)-C(n+2)]/K,* =3[T,(n)-TI(n+2)]/K,-3Q,(a0, ~0)

+ $[(9n + WQJO, 0)

- (4n +4)&&O, 2) +(n + l)Q,(G 2)1

-. $(n + l)[Q,,@, a0) - QJ-2, WI, _ [C(n)-C(n+2)]/K,b$- 1 +M6(3~)~(2~)1-~9/(4a)]

1 3-6 =--- 2 7F’

Since $ - (3 -&Q/W is positive (about 0. l), C(n) > C(n +2) for all large n. U

To approach the proofs of Lemmas 8 through 10 let the n = 2r voters in the conditioning event in the definition of Q,(tl, tz) be partitioned into I two-voter groups such that, within each group, one voter prefers 1 to 2 and the other prefers 2 to 1. For the ith two-voter group let ;~ci equal 1, 0, or -1 respectively according to whether both voters prefer 1 to 3, one prefers 1 to 3 and the other prefers 3 to 1, or both prefer 3 to 1; let yi equal 1, 0, or -1 respectively according to whether both voters prefer 2 to 3, one prefers 2 to 3 and the other prefers 3 to 2, or both prefer 3 to 2. Seven of the nine (xi, yi) pairs are possible-only (1, -1) and (-1,l) are excluded. The probability for each pair is shown in the following matrix, followed in parentheses by the number of groups that have that pair.

Y

1 0 -3

1 $(a,, &lp) 0

x 0 $(a,) $(a,) 6(aJ

-1 0 8(a5) &(a~)

For example, a1 groups have (4, yi) = (1, l), . . . , an3 a7 groups have (x+, yi) = (0, 0), with a, + b l l + a, = r. Nate dso tti%t t1 = 2[a, + a2- as - a6] and t2 = 2[a, + a3- a+- a,].

Proof of Lemma 8. The analysis of the preceding paragraph random vector (x,, yi) has mean (0, 0), covariance matrix

implies that each

and correlation coefficient p = $ between the two variables. Since (c X-J&, c y,/h) has a bivariate normal distribution with mean (0,O) and p = i as n --9 00, standard

, I tables (e.g., Yamauti [13n show that its nonnegative orthant probability ap-

I proaches 4 as PI+=. Since t+Qc*Cx+O, and t2~O*-*Cyi~O, it follows that I I Q,(sO, *O) -4. 0

Prdof 9. The lattice distribution treatment for the central limit theorem in Bhattacharya and Ranga Rao [2] or Bhattacharya [1, Theorem 2.11 implies that, for fixed @, fi),

where &, is the bivariate normal density with mean (C, 0) and covariance matrix V as given in the preceding proof. Since

&(O, 0) = [determinant ( V-‘)31nl(27t) = JF7P(W,

it follows that IIQ,,(c,, EJ -3&/(21r). 0

proof of L~~QMB 10. Lemzmas 5 and 6 show that r[Qzr(0)- Q2r(-2)]--0. she Q2St) = Q& 313) + QzJ;t, CO), ad s&e r[Q2,(2,0) - Q2,(0, O)]- 0 according to Lemma 9,

0 - ~CQ2,(O) - Q&2)]

= r(l Q,(O, 30) - Qd-2,30)] - [ Q2J -2, co) - a,,(O, co)-J

- rQ: Q2,(0,~0) - Q2,(-2, a@]- [Q2,(2, ~0) - a,,(O, >o)]

-1Q2A290) - Q2,(0, OHI

= 4Q2r(O, a0) - Qd-2, ~0)] - r[ Q2,(2, B 0) - a,,(O, ao)].

Therefore

6Q2r(Os aO)- Q2,(-2, bO)] h r[Q2,(2, 30) - Q2J0, a~)].

Then. since Q&, aO)- Q&i, 30)-O for fixed tl and t;,

(n + 1)[QJO, W- QJ-2, )-(I)]

- n[Q,(O, W- a,(-2, BO,]

= 2rIQd0, a0) - Q2,(-2, so)]

- r[Q2,(0, a0) - Qzr(-2, SO;]+ r[Q2r(2, 20) - a,,(O, >-0)]

= 4Q2,(2, ~0) - Q2r (-2, >O)]

-(r + 1)[Q2,(2, *O) -- Q2,(-2, SO)].

Therefore

(n + l)[Q,,(O, 20) - Qn(--2, ,-%I- (r + l)[u,,(Z~O) - Q&2, aW].

We shah now show that (r-p l)[&(2,80) - Q&2, BO)]S 9/(4m), which estab- lishes Lemma 10.

With the a, as defined in the matrix prior to the proof of Lemma 8,

where AI = (nonnegative integral (aI, u& c u = r} and A,={nonnegative

&+a3 aQd+Q(j, C & =I)* Let

. . . . u,): u,+uz=u5+u6+1, t2l+t23%44+

integral (uI,. . . , UT): uI+uz=uS+~-l,

A3= (~~,...,u,):u,+u,= I

u5+u6, u+0, u+0, ul+u+uq+&j,

u3+u(j<u,+u4, z a, =r+f

A5 = {ib: dr+i,r+l-b is even},

where all varialbles are nonnegative integers.

(r + l)[:Q3(2,W - Q&2, ao)]

- c (r+ l)![u$-(uz+ 1)]3”7 -

A2,U{as.= “l} n bi !)(Q* + 119

We shah prove that

= (r-l- l)!(a, - 43”’

A3 n (ai w

(I+ l)!kSb

dt2!t+j!(d++ k)!(as,+ k)!b!g’

c $(t + 1 - b)(r + l)!Sb = As (r+I-b)!b!9’

(7)

Three auxiliary lemmas will be used in verifying the We shall prove these before continuing with the [Qz,(2, aa) - Q&2, aO)ls 91(47r).

preceding sequence. proof that (r + 1)

Lenumt 11. Let B be the set of all nonnegative imegrul (n,, n2, n,) for wW2

(2)

(3)

(4)

(5)

(6)

RI -1- n2+ n3 = n, where n is a nonnegatitw integer. Then

c n3 in 2n

B nl!&!(nl+ n3)!(?@ n3)! =(n!)z n ( I) *

Lemma 12. *) < 2*V/G for every n E {1,2, . . .}.

Le3mmasr 13. Let E3, be the set of aU nonnegative ewn integers that do not exceed n, and let A be strictly between 0 ad 1. Then

Roof of Lemma Il. The identity

where the sum is over all nonnegative integral (x, y) such that x +y is constant, will be used in going from line three to line four in the ensuing sequence. The foregoing identity is noted, for example, by Feller [4, p. 621.

Proof of Lenlmrr 12. This follows easily from the bounds on Stirling’s approxima- tion for n! given, for example, on p. 52 in Feller [4]. Cl

Proof of Lemnra 13. Let p(n) be the noted sum over E, in Lemma 13. Then

P(n+l)=(l-h)p(n)+h[l--p(n)].

248 P.C. FWabum, W.V. G&&h, E. Maskin

Figuratively speaking, the probability of an even number of successes in n + 1 Bernoulli trials with success probability A equals the probability of an even number of successes in the first n trials times the probability of failure at trial y1+ 1, plus the probability of an odd number of successes in the first n trials times the probability of success at trial II + 1. Since O< h < , p(n) approacires the stable

p that is the solution to p=(l-A)p+h(l-p), which is p=i. Cl

We now return to sequence (2) through (7). Given a = (a,, . . . , a?), let a’ and a* be defined by

a;l=a,+l, a;=aS- 1, u; = q for jg!{2r 5);

aT=a,, a: = a,; t-44: = a,, a4 = a2; a: =q for jE{3,4,7}.

To go from (r + 1)[Q2,(2, 30) - Q2J-2, aO)] to (2), note first that all vectors In A1 are uniquely generated from the vectors in A2 that have as >O by the mapping a + Q’ except for those that have ai = 0. However, since

s (I + l)(&

and since (r + l)(g)’ - 0 by l’Ho$pital’s rule, the omission from Al of vectors that have a2 = 0 will not affect the large r behavior of (r + 1)Q2,(2, 80). In like manner the 15= 0 vectors can be omitted from (r + l)Q2,1-2, 20) so that

which verifies (2). To obtain (3) from (2), first replace a,+ 1 in (2) by a: = a2+ 1, then drop the

double prime to get (2) equal to

x (I+ l)!(lZ, - l&2)3’

&o n bi w

with

A30={(al, l l l , a,): a,+a2:=as+a6, a,+a3aa4+u6,

u,>o, a,>o, L @ =r+l}.

Given 2 c A3,,, the mapping Q -+ a* gives a* E ABO if and only if a.,+ a3 2 a,+ aI, and if a, a* E As0, then their terms in the A30 sum cancel since at -a$ =

4% -ad. After all such cancellations, A 30 retains only those Q for which a6+ aa< ad+ al, and it is these vectors that comprise A,. Hence (3) equals (2).

The impdiM al+a+a4+& md a3+~<a1+a4 in AS imply al>u6. Let

k=a,-G. Since as-~=al-(16 in At, (3) equals

Z (r+l)Ik3”1

As, as!%!&+ &)!(a,+ k)!a3!a&!y

with

AJ1 = {(az, ab, k as,a,,a7): k>0,az>0,a,-aa36k,as-a,ek,

2(az+e+k)+(a3+a4+a7)=r+l}.

The sum of 3%/(a3!a4!a7!) over all (Up, a4, a,) for which u3+a4+u7 = b is Sb/b!. Since this sum ignores the inequality constraints in Aj1, and since A4 is just As1 minus its inequality constraints and with b = a3 + a4+ a7, it follows that (3) is less than (4).

In A4, b is feasible if and only if 6 B 0 and $(I + 1 - b) is an integer, in which case t + 1 -b must be even. With b feasible and fixed, we next sum the terms in

that involve k, a2 and a6 over d ( a2, a& k) for which a2 + a6 + k = $( r + 1 - b). Lemma 11, this sum is

k

When this is taken account of in (4), it follows that (4) equals (5). The term for b=r+l or r+l-b=O in (5) equals zero. In addition, when

r+l-b is positive and even, Lemma 12 gives

When the right-hand side of this inequality is substituted left-hand side, we obtain (6) so that (5) is less than (6).

Finally, Lemma 13 implies that

into (5) in place of its

and therefore (6) -9/(4?r). The sequence of (2) through (7) thus shows that (r-k l)[Q*,(2, SO)-Q,,(-2, zO)]S g/(4=), and the proof of Lemma 10 is com- plete.

250 P.C. Fishbum, W.V. CM&in, E. Maskin

6. Additiod resulti for odd n

It seems apparent from preceding sections that Kelly’s conjectures are more amenable to resolution for odd n than for even n. Therefore, in attemptiag to *

extend the results derived above, it would appear more profitable to work initially with odd n. The main results we have obtained for odd II are Theorem 1, which says that

C( m, 3) > C(m + 1,3) for all m E {2,3, . . .}, (8)

and Theorem 2, which says that

~(3, n) > C(3, n +2) for all odd n 3 1. (9)

In concluding this study vve note several related results for odd n that arise from the following recursion relations in Gehrlein and Fishbum [6]:

C(4, n) = 2C(3, n) - 1 for odd n s 1

C(6, n) = 3 -X(3, n)+3C(S, n) for odd n & 1.

(10)

(11)

Theorem 6 follows the spirit of (8), while Theohb,m 7 is in the mode of (9).

Theorem 6. T1ze following hold for all odd n a 3: (a) C(3, n)> C(4, n): (b) C(3, n) ’ C(5, n); (c) C(5, n) > C(6, n) + C(4, n) > C(5, n): (d) C(3, n) > C(6, n) if and only if C(4, n) > C(5, n).

Proof. (a): By (lo), C(4, n)-C(3, n) = C(3, n)- 1 CO. (b): If C(5, n) 2 C(3, n), then (11) implies that C(6, n) 3 1+2[1- C(5, n)],

which is false since 1+2[1- C(5, n)]> 1. (c): By (k1), C(?, n) = 3 -4C(3, n) + 3C(5, n)-C(69 n). Since

l> C(3, n), -2 + 4C(3, n) - 2C(5, n) ) C(5, n)- C(6, n).

But, by (lo), -2 + 4C(3, n) = 2C(4, n). Therefore

2[C(4, n) - C(5, n)]> C(5, n) - C(6, n).

(d): By (10) and (ll), C(6, n)- C(3, n) = 3[C(5, n) - C(4, n)]. 0

Theorem 7. The following hold for all odd n 3 1: (a) C(4, nPC(4, n+2); (b) C(6, n) > Ci6, n + 2) + C(5, n) -a C(5, n + 2).

Pro& (a): By (IO), C(4, n)-C(4, n +2) = 2[C(3, n)- C(3, n + 2)]. Then, by (9), C(4, n) ) C(4, n + 2).

(W: By (10,

3[C(5, n)-C(5, n+2)]=[C(6, n)-C(6, n+2)]+5[C(3, n)-C(3, n+2)].

Hence (9) and C(6, n)> C(6, n +2) imply that C(5, n)> C(5, n +2). 0

Theorems 1 and 6 imply that if C(5, n) > C(6, n) is true for all odd n a 3, then C(3, n) > C(4, n)> C(5, n) > C(6, n) for all odd n a3. Hence a reasonable next step for Conjecture 1 would be to try to establish C(5, n)> C(6, n) for odd II 2 3. In like manner, it follows from Theorems 2 and 7 that if C(6, n) > C(6, n +2) for odd nal, then C(m,n)>C(m,n+2) for odd nal and otr~{3,4,5,6}. Hence a reasonable next step for Conjecture 2 wouId be to try to estabkh C(6, n) 3 C(6, n+2) for odd n * 1. Moreover, If this is true, then-as shown by our final theorem-it is true also that C(5, n)> C(6, n) for odd n zz 3.

Tkorern 8. If C(6, n) > C(6, n +2) for odd n a 3, then C(5, n) > C(6, n) for odd na3.

pnrof, Assume that C(6 n) > C(6, n +2) for odd n 2 3 and, contrary to the theorem, suppose that C(6, n)a C(5, n) for some odd n a 3. Then, by (1 l), 2C(5, n) ss SC(3, n)-3. By (9) and GuSbaud’s result [S, 9]* C(3, n) >0.912, and therefore C(6, n)B C(5, n) implies that C(5, n) >0.78. In order for this to be true, Theorem 7(b), the hypothesis that C(6, n) > C(6, n +2), and known values [6] of C(5, n) require n s 7. But values [6] of C(5, n) and C(6, n) give C(5, n) > C(6, n) for n ~{3,5,7}. Ll

R.N. Bhattacharya, Refhements of the muhidimensionaI central limit theorem and applications, Ann. Probability 5 (1977) l-27. R.N. Bhattacharya and R. Ranga Rao, Normai Approximations and Asymptotic Expansions (Wiley, New York, 1976). F. DeMeyet and CR. Plott, The probability of a cycIicaI majoriry, Econometrica 38 (t 970) 345-354. W. Feller, An Introduction to probability Theory and Its Applications, Vol. 1, 2nd edn (Wiley, New York, 1957). M. Garman and M. Kamien, The paradox of voting: Probability calculations, Behavioral Sci. 13 (1968) 306316. W.V. GeMein and P.C. Fishburn, The probability of the paradox of voting: A computable solution, J. Econ. Theory 13 (1976) 14-25. W.V. Gehrlein and P.C. Fishbum, Proportions of profiles with a majority candidate, Cornput. and Math. with Appl., 5 (1979) 117-124. W.V. Gehrlein and P. C. Fishbum, Probabilities of electron outcomes for large electorates, J. Econ. Theory 19 (1978) 38-49.

252 P.C. Fishbum, W.V. Gehrfein, E. Ma&n

[9] G.T. Guilbaud, Les th&ories de I’int&& g&n&al et le probl&me logique de l’agr&gation, &on. Appl. 5 (1952) 501-584.

[lo] J.S. Kelly, Voting anomalies, the number of voters, and the number of alternatives, Economet- rica 42 (1974) 239-251.

[ 1lJ R.M. May, Some mathematical remarks on the paradox of voting, Behavioral Sci. 16 (1971) 143-151.

[I23 R.G. Niemi and H.F. Weisberg, A mathematical solution to the probability of the paradox of voting, Behavioral Sci. 13 (1968) 317-323.

Q33 2. Yamauti, ed., Statistical Tables and Formulas with Computer Applications (Japanese Stan- dards Association, Tokyo, 1972).